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Find the Equation of the Plane Passing Through the Intersection of the Planes → R . ( ˆ I + ˆ J + ˆ K ) = 1 and → R . ( 2 ˆ I + 3 ˆ J − ˆ K ) + 4 = 0 and Parallel to X-axis. - Mathematics

Sum

Find the equation of the plane passing through the intersection of the planes `vec(r) .(hat(i) + hat(j) + hat(k)) = 1"and" vec(r) . (2 hat(i) + 3hat(j) - hat(k)) +4 = 0 `and parallel to x-axis. Hence, find the distance of the plane from x-axis.

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Solution

`vec(r) . (2hat(i) + 3hat(j) - hat(j) ) = 1 " and " vec(r) . (2hat(i) + 3 hat (j) -hat(k)) + 4 = 0`

`vec(r) . (hat(i) + hat(j) - hat(j) ) = 1      - ( vec(r) . (2hat(i) + 3 hat (j) -hat(k))= 4` 

Taking ` r = xhat(i) + yhat(j) - zhat(k)`

x +y +z = 1    -2x - 3y + z = 4

Equation of plane is L1 + λL2 = 0 

(x + y +z - 1) + λ (-2x -3y + z -4) = 0

= (1 -2 λ)x + (1 - 3 λ) + (1 + λ ) z+ (-1 -4 λ) = 0

As given it is parallel to x-axis.
⇒ Normal of the plane is perpendicular to x-axis
Direction Ratio’s of `(1 -2λ) hat(i) + (1 -3λ)hat(j) + (1 + λ) hat(k) `

                                1 - 2λ, 1-3 λ ,1 + λ

Direction ratios of x-axis are 1, 0, 0
So, (1- 2λ) ×1 + (1- 3λ) ×0 + (1+λ )×0 = 0

⇒` lambda =1/2`

⇒ Equation of plane is

`(1-2(1/2)x + (1-3/3)y +(1+1/2)z(-1-4(1/2)))= 0`

ox + `((-1)/2 ) y + (3/2) k - 3 = 0`

⇒ ox + y - 3z + 6 = 0
⇒ y - 3z + 6 = 0 
Distance of the plane from x-axis is

`lambda = (|6-0|)/(sqrt(1^2 +3^2 +0^2)) = 6/sqrt(1+9) = 6/sqrt(10)`

`lambda = 6/sqrt(10)`

Concept: Vector and Cartesian Equation of a Plane
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